CBSE 2026 · Region 4 · Set 1 · Q32 · 5 marks
Find : \[\int \frac{\mathrm{x}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+4\right)} \mathrm{dx} \]Evaluate : \[\int_{0}^{1} \frac{\mathrm{x} \tan ^{-1} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{3 / 2}} d \mathrm{x} \]
Find : \[\int \frac{\mathrm{x}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+4\right)} \mathrm{dx} \]
Evaluate : \[\int_{0}^{1} \frac{\mathrm{x} \tan ^{-1} \mathrm{x}}{\left(1+\mathrm{x}^{2}\right)^{3 / 2}} d \mathrm{x} \]
Marking-scheme solution
Let $\displaystyle \dfrac{x}{(x-1)\left(x^{2}+4\right)}=\dfrac{A}{x-1}+\dfrac{B x+C}{x^{2}+4}$
on solving we get, $\displaystyle A=\dfrac{1}{5}, B=\dfrac{-1}{5}, C=\dfrac{4}{5}$
so, $\displaystyle \int \dfrac{x}{(x-1)\left(x^{2}+4\right)} d x=\dfrac{1}{5} \int \dfrac{d x}{x-1}-\dfrac{1}{10} \int \dfrac{2 x d x}{x^{2}+4}+\dfrac{4}{5} \int \dfrac{d x}{x^{2}+4}$
$\displaystyle \int \dfrac{x}{(x-1)\left(x^{2}+4\right)} d x=\dfrac{1}{5} \log |x-1|-\dfrac{1}{10} \log \left(x^{2}+4\right)+\dfrac{2}{5} \tan^{-1}\left(\dfrac{x}{2}\right)+c$
Let $\displaystyle x=\tan t \Rightarrow d x=\sec^{2} t\, d t$
$\displaystyle I$ (Let) $\displaystyle =\int_{0}^{1} \dfrac{x \tan^{-1} x}{\left(1+x^{2}\right)^{\frac{3}{2}}} d x=\int_{0}^{\frac{\pi}{4}} \dfrac{t \tan t}{\sec^{3} t} \sec^{2} t\, d t=\int_{0}^{\frac{\pi}{4}} t \sin t\, d t$
$\displaystyle I=(-t \cos t+\sin t)_{0}^{\frac{\pi}{4}}$
$\displaystyle I=\dfrac{-\pi}{4 \sqrt{2}}+\dfrac{1}{\sqrt{2}}$ or $\displaystyle \dfrac{-\pi+4}{4 \sqrt{2}}$
IntegralsIntegration by Partial FractionsApplylong_answerhard
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.